Problem: In the right triangle shown, $BC = 3$ and $AB = 10$. What is $AC$ ? $A$ $C$ $B$ $?$ $3$ $10$
We know $a^2 + b^2 = c^2$ We want to find $b$ ; let $a = 3$ and $c = 10$ So $b^2 = c^2 - a^2 = 10^2 - 3^2 = 91$ Then, $b = \sqrt{91}$.